Class 13

Chapter 4 Objectives

• Identify and describe the standard normal and normal distributions

• Standardize normal distributions and calculate Z-scores

• Calculate percentiles and exact probabilities

• Apply the 68-95-99.7 Rule

• Read a QQ-Plot (not in book)

The Normal Distribution

Notation

• \(\mu\) (Greek letter mu) represents the mean

• \(\sigma\) (Greek letter sigma) represents the standard deviation of the mean

• \(N(\mu, \sigma)\) stands for a normal distribution with mean \(\mu\) and standard deviation \(\sigma\)

Histograms and Density Curves

Vocabulary scores for 947 seventh-graders. Both histograms and density curves can be helpful in identifying normal distributions.

Example: OkCupid, Heights of Males

Vocabulary scores for 947 seventh-graders. Both histograms and density curves can be helpful in identifying normal distributions.

Example: OkCupid, Heights of Females

Dashed line is self-reported heights by females on OkCupid. Dark purple line is the normal distribution with the same mean and standard deviation. Light purple line is the US average.

The shape of a normal distribution varies by location (mean) and scale (standard deviation)

Changing the mean shifts the “center” of the distribution. Changing the standard deviation alters the “width” of the distribution (i.e. variability).

Standardizing Normal Distributions with Z-Scores

• A Z-score is the number of standard deviations a value falls above (when positive) or below (when negative) the mean of the data

• Z-scores standardize a normal distribution by…

  • Centering the data at 0 by subtracting the mean from each score

  • Scaling the units of the data to 1 by dividing the centered data by the standard deviation

Calculating the Z-Score

\[ \begin{aligned} Z&=\frac{\operatorname{observed value}-\operatorname{mean}}{\operatorname{standard deviation}} \\ &= \frac{x-\mu}{\sigma} \end{aligned} \]

What does “centering” the data mean?

• The numerator of the Z-Score \(x-\mu\) calculates how many units an observed value is from the mean of the normal distribution

• When \(x_i \approx \mu\), \(x_{\operatorname{centered}} \approx 0\)

• The units of random variable \(X_{\operatorname{centered}}\) are the same as the units for the original variable \(X\)

Properties of Centered Normal Distributions

For a given random variable \(X\) with a normal distribution, you center the data by calculating \(x_i-\mu\) for each value of \(X\) such that…

• When \(x_i > \mu\), \(x_{\operatorname{centered}}>0\) and is interpreted “\(x_{\operatorname{centered}}\) is \(x_i-\mu\) units greater than the mean \(\mu\)”
• When \(x_i = \mu\), \(x_{\operatorname{centered}}=0\)
• When \(x_i < \mu\), \(x_{\operatorname{centered}}<0\) and is interpreted “\(x_{\operatorname{centered}}\) is \(\mu-x_i\) units less than the mean \(\mu\)”

What does “scaling” the data mean?

• Dividing the numerator of the Z-score \(x-\mu\) by the denominator \(\sigma\) converts the units of the centered data to standard deviations

• Converts “\(x_{\operatorname{centered}}\) is \(x_i - \mu\) units greater/lesser than the mean \(\mu\)” to “\(x_{\operatorname{scaled}}\) is \(x_i - \mu\) units greater/lesser than the mean

• When \(x_i-\mu \approx \sigma\), \(x_{\operatorname{centered}} \approx 1\)

• For scaled data, \(1 \operatorname{unit} = 1 \operatorname{standard deviation}\)

Example: Test Scores

• SAT scores are normally distributed with \(\mu=1500\) and \(\sigma=300\) (\(N(\mu=1500, \sigma = 300)\))

• ACT scores are normally distributed with \(\mu=21\) and \(\sigma=5\) (\(N(21, 5)\))

How do we compare normal distributions with different locations and scales? Is Pam more above average than Jim? Vice versa?

Zooming in

If both Pam and Jim applied to John Carroll, who would be the better student to admit?

Example: Calculating Pam’s SAT Z-Score

If SAT scores have the distribution \(N(\mu=1500, \sigma=300)\) and Pam’s SAT score is 1800, then Pam’s Z-score is…

\[ \begin{aligned} \operatorname{Z-Score}&=\frac{x-\mu}{\sigma} \\ &= \frac{1800-1500}{300} \\ &= 1 \end{aligned} \]

Pam’s SAT Z-score is 1, so Pam’s SAT score is 1 standard deviation greater than the mean.

Example: Pam’s SAT Z-Score in R

pam_mean <- 1500
pam_sd <- 300

pam_centered <- 1800 - pam_mean

print(pam_centered)

[1] 300

pam_z <- pam_centered / pam_sd

print(pam_z)

[1] 1

Example: Pam’s SAT Z-Score in R

(1800 - 1500) / 300

[1] 1

Example: Calculating Jim’s ACT Z-Score

If ACT scores have the distribution \(N(\mu=21, \sigma=5)\) and Jim’s ACT score is 24, then Jim’s Z-score is…

\[ \begin{aligned} \operatorname{Z-Score}&=\frac{x-\mu}{\sigma} \\ &= \frac{24-21}{5} \\ &= 0.6 \end{aligned} \]

Jim’s ACT Z-score is 0.6, so Jim’s ACT score is 0.6 standard deviations greater than the mean.

Putting it together…

The Standard Normal Distribution

• The standard normal distribution is a normal distribution with \(\mu=0\) (centered) and \(\sigma=1\) (scaled)

• The standard normal distribution is written \(N(\mu=0, \sigma=1)\))

• Units of the standard normal distribution are standard deviations (Z-scores) (i.e. 1 unit = 1 SD)

• Observations that are 2+ standard deviations from the mean are considered unusual

The 68-95-99.7 Rule

When data is (nearly) normally distributed…

• ~68% of the observations are within 1 standard deviation of the mean (\(\mu \pm \sigma\))

• ~95% of the observations are within 2 standard deviations of the mean (\(\mu \pm 2\sigma\))

• 99.7% of the observations are within 3 standard deviations of the mean (\(\mu \pm 3\sigma\))

The 68-95-99.7 Rule

The 68-95-99.7 Rule describes approximately what proportion of the observations should lie within 1, 2, and 3 standard deviations of the mean respectively, if the data is normally distributed

Example: Test Scores

Proportions, Probabilities, and Percentiles

A percentile is the proportion or percentage of observations that fall below a given threshold in a distribution.

Proportions, Probabilities, and Percentiles

\[ \begin{aligned} \operatorname{Percentile}(X=x_i) &= \frac{\operatorname{count}(\operatorname{observations} \le x_i)}{\operatorname{count}(\operatorname{total observations})} \\ &=\operatorname{Proportion}(\operatorname{observations} \le x_i) \\ &=\operatorname{Probability}(\operatorname{any observation} \le x_i) \end{aligned} \]

Probability Density and Cumulative Density Functions

Probability Density Function for Normal Distributions

You too can calculate probabilities for continuous numeric variables!

\[ P(X=x_i)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x_i-\mu)^2}{2\sigma^2}} \]

Calculating Percentiles with Z-Score Tables

You can use a Z-Score Table to look up the percentile that corresponds to a particular Z-Score for a standard normal distribution.

Calculating Probabilities with Z-Score Tables

You can use a Z-Score Table to look up the probability that an observed Z-Score is less than or equal to a given Z-Score (i.e. threshold) for a standard normal distribution.

Calculating Percentiles in R

# Parameter 1 = value to look up
# Pam's SAT Z-Score --> Percentile
pnorm(1, mean = 0, sd = 1)

[1] 0.8413447

# Parameter 1 = value to look up
# Jim's ACT Z-Score --> Percentile
pnorm(0.6, mean = 0, sd = 1)

[1] 0.7257469

What if we want to know the percent ABOVE a threshold?

Probabilities above a threshold

• Area under a probability curve = 1 (i.e. sample space)

• Probability above a threshold = 1 - percentile of threshold

• \(P(X \le x_i)=1-P(X \not \le x_i)\)

• \(P(X > x_i)=1-P(X \le X_i)\)

Easy to find the probability of the complement in R

pnorm(1, mean = 0, sd = 1, lower.tail = F)

[1] 0.1586553

1 - pnorm(1, mean = 0, sd = 1)

[1] 0.1586553

Other: Discrete Numeric Variables

Sometimes the normal distribution is an acceptable approximation of a discrete numeric variable, but other distributions may be more appropriate.

Other: QQ-Plot

Quantile-Quantile (QQ) Plots can help easily identify when you can and cannot assume normality.